JavaScript:冷凝if else声明

Question

是否有更短的更有效的方法来做到这一点？它似乎有点沉重,我只是想知道它是否可以浓缩？

 var y = []

  for(let i=0;i < word.length;++i){
    if(word[i] == "A"|| word[i] == "a"){
      y.push(0)
    }
    else if(word[i] == "B"|| word[i] == "b"){
      y.push(1);
    }
    else if(word[i] == "C"|| word[i] == "c"){
      y.push(2);
    }
    else if(word[i] == "D"|| word[i] == "d"){
      y.push(3);
    }
and so on..


  return(y);
}

Answer 1

一种选择是使用一个字符数组,然后.indexOf用来查找字符的索引:

const word = 'bBac';
const chars = ['a', 'b', 'c', 'd'];

const y = [...word].map(char => chars.indexOf(char.toLowerCase()))
console.log(y);
// return y;

为了更好的效率,而不是.indexOf(即O(N)),使用a Map(O(1)):

const word = 'bBac';
const charMap = new Map([
  ['a', 0],
  ['b', 1],
  ['c', 2],
  ['d', 3]
]);

const y = [...word].map(char => charMap.get(char.toLowerCase()))
console.log(y);
// return y;